OCR A-Level Chemistry Periodic table, elements and physical chemistry (01), June 2024: Question 17
7 marks · Hard difficulty · Extended Response
Explain why the gradient decreases over time using collision theory, determine the order of reaction with respect to crystal violet, find the rate of reaction at three minutes, and calculate the rate constant.
Practise this questionQuestion
Question text
17 Crystal violet (CV) is a purple dye. In the presence of an alkali, CV reacts to form a colourless
product.
A student uses a colorimeter to investigate the rate of the reaction between CV and sodium
hydroxide, NaOH.
• The student mixes 10.0 cm3 of 2.8 × 10–7 mol dm–3 CV with 10.0 cm3 of 0.016 mol dm–3
NaOH.
• A large excess of NaOH is used, so that the reaction is effectively zero-order with respect to
OH– ions.
• The student places a sample of the reaction mixture in a colorimeter and measures the
absorbance over time.
The student uses the absorbance readings to calculate the concentration of CV and plots a graph
of concentration of CV against time, as shown below.
1.6 × 10–7
1.4 × 10–7
1.2 × 10–7
1.0 × 10–7
oncentration
f CV 0.8 ×10–7
mol dm–3
0.6 × 10–7
0.4 × 10–7
0.2 × 10–7
01 2 3 4 5 6 7 8 9 10
time/min
(a) Using collision theory, explain why the gradient decreases over time.
… [1]
(b)* Use the graph to determine the order of reaction with respect to CV, the rate of the reaction at
three minutes and the rate constant, k.
Your answer must show full working on the graph and on the lines below. [6]
Extra answer space if required.
Mark scheme
Show the mark scheme
Question Answer Marks Guidance
17 (a) (Over time) concentration decreases AND collisions 1 ALLOW less moles/particles per unit volume.
are less frequent ✓ ALLOW fewer collisions per second/per unit time
IGNORE (over time) fewer reacting particles
IGNORE …chance of..
IGNORE amount decreases
IGNORE successful
IGNORE particles more spread out/further apart
DO NOT ALLOW particles have less energy in terms of
energy distribution.
(b)* Please refer to the marking instructions on page 4 of this 6 Indicative scientific points may include:
mark scheme for guidance on how to mark this question.
Level 3 (5–6 marks) Care: ALLOW the use of ECF for values obtained from a previously,
A comprehensive conclusion using quantitative data incorrectly, calculated value.
from graph to correctly determine
1st order conclusion for CV using half lives/gradients ALLOW minor slips as we are looking for a holistic approach
AND rate at 3 minutes to LoR marking.
AND determination of k
Minutes Seconds
There is a well-developed line of reasoning which is Half life values 2.4 to 2.6 min 144 to 156 s
clear and logically structured.
Rate at three (-) (1.5 to 1.8) ×10–8 (-) (2.5 to 3.0)×10–10
minutes mol dm–3 min–1 mol dm–3 s–1
Level 2 (3–4 marks)
Reaches a conclusion using quantitative data from –1 –3 –1
graph to correctly determine rate at 3 minutes AND Value of k 0.24 to 0.30 min (4.0 to 5.0) x 10 s
determination of k. –1 –1
Units of k min s
OR
Half- lives/gradient with 1st order conclusion for CV
AND determination of k
OR
determined rate AND half-life/first order for CV Examples of the communication statement being met would
OR typically include:
Attempts to determine rate, k and order for CV
• For L1 and L2: full working on the graph and/or
There is a line of reasoning with some structure and appropriate units for calculated values.
supported by some evidence.
• For L3: full working on the graph and appropriate units
Level 1 (1–2 marks) for calculated values.
Reaches a simple conclusion using at least one piece
of quantitative data from the graph, i.e.
Attempts to calculate rate at three minutes OR k OR
links half lives to 1st order.3
There is an attempt at a logical structure with a
reasoned conclusion from the evidence.
0 marks No response worthy of credit If time has been measured in minutes
(see below for values using seconds).
Indicative scientific points may include:
Evidence for 1st order
1st order clearly linked to half-life OR 2 gradients:
Half life
Half- life shown on graph
Half- life range 2.4 to 2.6 min
Two ‘constant’ half lives
OR Two gradients → two rates
2 tangents shown on graph at c and c/2
This could include c = 0.61 × 10–7 mol dm–3 (t = 3 min)
Gradient at c/2 is half gradient at c
e.g. c = 0.8 x 10–7 mol dm–3, gradient = 2.2 × 10–8 (mol dm–3
min–1)
AND c = 0.4 x 10–7 mol dm–3, gradient = 1.1 × 10–8 (mol dm–3
min–1)
For chosen method, conclude that the reaction is 1st order wrt CV.
Rate at three minutes
Tangent shown on graph as line at t = 3 min
Gradient in range: (1.5 – 1.8) × 10–8
rate as gradient with units: mol dm–3 min–1
ln2 –1
OR k = t½ = 0.28min
And k substituted into rate equation.
e.g.
Rate = k [CV]
Rate = 0.277 x 0.61 x10-7
= 1.7 x10-8 mol dm–3 min–1
Determination of k
k clearly linked to rate OR half-life:
rate 1.75 x 10–8
e.g. k = [CV] = 0.62 x 10–7 = 0.28
k in range: 0.24 - 0.30 min–1
ln2 –1
OR e.g. k = t½ = 0.28 min
Units of k: min–1
If time has been measured in seconds:
Evidence for 1st order
1st order clearly linked to half-life OR 2 gradients:
Half life
Half- life shown on graph
Half- life range 144 to 156 s
Two ‘constant’ half lives
OR Two gradients → two rates
2 tangents shown on graph at c and c/2
This could include c = 0.6 × 10–8 mol dm–3 (t = 3 min)
Gradient at c/2 is half gradient at c
e.g. c = 0.8 x 10–7 mol dm–3,
gradient = 3.7 × 10–10 mol dm–3 s–1
AND c = 0.4 x 10–7 mol dm–3,
gradient = 1.8 × 10–10 mol dm–3 s–1
For chosen method, conclude that the reaction is 1st order wrt CV.
Rate at 180 seconds
Gradient in range (2.5 to 3.0) ×10–10
rate as gradient with units: mol dm–3 s–1
ln2 –3 –1
OR k = t½ = 4.6 x 10 s
And k substituted into rate equation.
e.g.
Rate = k [CV]
Rate = 0.00462 x 0.61 x10-7
= 2.8 x 10-10 mol dm–3 s–1
Determination of k
k clearly linked to rate OR half-life:
17 rate 2.75 x 10–10
e.g. k = = = 4.4 x 10–3 s–1
[CV] 0.62 x 10–7
k in range (4.0 to 4.8) x 10–3 s–1
ln2 –1 –3 –1
OR e.g. k = t½ = 0.28 min OR 4.6 x 10 s
Units of k: s–1
How to answer it
Kinetics: Colorimetry & Rate Calculations
What this question tests
This question assesses your understanding of chemical kinetics using graphical methods. You are tested on applying collision theory to rate changes, determining order of reaction using concentration-time graphs (via half-lives or gradients), calculating instantaneous rates using tangents, and determining the rate constant (k) with appropriate units.
Explaining Rate Decreases Over Time
✅ Correct Answer
Over time, concentration decreases AND collisions are less frequent.
💡 Key Knowledge
- Reaction rate depends on the frequency of successful collisions.
- As reactants are consumed, concentration decreases, meaning fewer particles occupy a given volume.
❌ Common Errors
- Saying particles have "less energy" (kinetic energy distribution does not change simply because concentration drops).
- Vague phrasing like "amount decreases" or "particles spread out". Keep it strictly to concentration and collision frequency.
Determining Order, Rate at 3 Minutes, and Rate Constant (k)
✅ Correct Answers & Level Descriptors
Order: 1st order with respect to CV (demonstrated by constant half-lives or decreasing gradients).
Rate at 3 min: Range 1.5 × 10⁻⁸ to 1.8 × 10⁻⁸ mol dm⁻³ min⁻¹ (if measured in minutes).
Rate Constant (k): Range 0.24 to 0.30 min⁻¹ (or 4.0 × 10⁻³ to 5.0 × 10⁻³ s⁻¹ ).
🧠 Exam Technique & Level of Response (LoR)
To achieve Level 3 (5–6 marks), you must show clear quantitative work on the graph itself (draw tangents or half-lives) and logical calculation steps on the answer lines with correct units.
📐 Step-by-Step Calculation Guide
- Determine Order: Show that successive half-lives are constant (e.g., time taken for concentration to drop from 1.4 × 10⁻⁷ to 0.7 × 10⁻⁷, then from 0.7 × 10⁻⁷ to 0.35 × 10⁻⁷ is roughly constant at ~2.5 minutes). Alternatively, draw two tangents at halves of a concentration to show rate halves. Conclude 1st order.
- Find Rate at 3 Minutes: Draw a sharp tangent to the curve exactly at t = 3 min . Calculate the gradient (change in y ÷ change in x). Ensure you factor in the scale axis multiplier ( ×10⁻⁷ ).
- Calculate k: Use either half-life formula k = ln(2) ÷ t½ (giving ~0.28 min⁻¹) OR rearrange the rate equation Rate = k[CV] to give k = Rate ÷ [CV] at a chosen point on the curve.
- Units for k: For a 1st order reaction, units are min⁻¹ (or s⁻¹ if time was converted to seconds).
❌ Common Calculation Traps
- Forgetting to include the ×10⁻⁷ power of ten when reading values off the y-axis.
- Drawing too-small triangles for tangents, which introduces significant reading inaccuracies.
- Omitting or giving incorrect units for the rate constant k .
Topics
Module 5: Physical chemistry and transition elements · Practical Activity Groups · PAG 9: Rates of reaction – continuous monitoring method · 5.1 Rates, equilibrium and pH
Question and mark scheme from the OCR A-Level Chemistry examination, Periodic table, elements and physical chemistry (01), June 2024. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.